Answered

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What is the product? StartFraction 4 k + 2 Over k squared minus 4 EndFraction times StartFraction k minus 2 Over 2 k + 1 EndFraction StartFraction 4 Over 2 k + 1 EndFraction StartFraction 2 Over k minus 2 EndFraction StartFraction 2 Over 2 k + 1 EndFraction StartFraction 2 Over k + 2 EndFraction

Sagot :

abidemiokin

Answer:

[tex]\frac{2}{k+2}[/tex]

Step-by-step explanation:

Given the expression:

[tex]\frac{4k+2}{k^2-4} * \frac{k-2}{2k+1}[/tex]

We are to find the product as shown:

[tex]= \frac{4k+2}{k^2-4} * \frac{k-2}{2k+1}\\\\= \frac{2(2k+1)}{k^2-2^2} * \frac{k-2}{2k+1}\\\\= \frac{2(2k+1)}{(k-2)(k+2)} * \frac{k-2}{2k+1}\\\\= \frac{2}{k+2} * 1\\\\= \frac{2}{k+2}[/tex]

Hence the required function is [tex]\frac{2}{k+2}[/tex]

d. StartFraction 2 Over k + 2 EndFraction

aka

d. 2/k+2

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